Capture-time Extremal Cop-Win Graphs

Transcription

1 Capture-time Extremal Cop-Win Graphs arxiv: v2 [math.co] Feb 208 David Offner Department of Mathematics and Computer Science Westminster College New Wilmington, PA, U.S.A. Kerry Ojakian Department of Mathematics and Computer Science Bronx Community College (CUNY) Bronx, NY, U.S.A July 6, 208 Abstract We investigate extremal graphs related to the game of cops and robbers. We focus on graphs where a single cop can catch the robber; such graphs are called cop-win. The capture time of a cop-win graph is the minimum number of moves the cop needs to capture the robber. We consider graphs that are extremal with respect to capture time, i.e. their capture time is as large as possible given their order. We give a new characterization of the set of extremal graphs. For our alternative approach we assign a rank to each vertex of a graph, and then study which configurations of ranks are possible. We partially determine which configurations are possible, enough to prove some further extremal results. We leave a full classification as an open question. Keywords: Pursuit-evasion games, Cops and robbers, Cop-win graphs, Extremal graphs Introduction The game of cops and robbers is a perfect-information two-player pursuit-evasion game played on a graph. To begin the game, the cop and robber each choose a vertex to occupy, with the cop choosing first. Play then alternates between the cop and the robber, with the cop moving first. On a turn a player may move to an adjacent vertex or stay still. If the cop and robber ever occupy the same vertex, the robber is caught and the cop wins. If the cop can force a win on a graph, we say the graph is cop-win. The game was introduced

2 by Nowakowski and Winkler [6], and Quilliot [0]. A nice introduction to the game and its many variants is found in the book by Bonato and Nowakowski []. One of the fundamental results about the game is a characterization of the cop-win graphs as those graphs which have a cop-win ordering [6], [0]. Independently, Clarke, Finbow, and MacGillivray [] and the authors of this paper [7] developed an alternative characterization that we call corner ranking. A thorough discussion of the similarities and differences of our approach is given in [7]. As with cop-win orderings, corner ranking characterizes which graphs are cop-win. Corner ranking can also be used to determine the capture time of a cop-win graph, as well as describe optimal strategies (in terms of capture time) for the cop and robber, where the capture time of a cop-win graph is the fewest number of moves the cop needs to guarantee a win, not counting her initial placement (for example, on the path with 5 vertices, the capture time is 2). In Section 2, we describe the corner ranking procedure and some useful properties of it that were proved in [7]. Bonato et. al. [2] make the following interesting definition. Definition.. Suppose n > 0 is a natural number. Let capt(n) = the capture time of a cop-win graph on n vertices with maximum capture time. For example, capt(4) = 2 since a path on four vertices has capture time 2, and no graph with 4 vertices has a capture time greater than 2. Define a cop-win graph with n vertices to be CT-maximal if no other cop-win graph on n vertices has a larger capture time. Building on [2], Gavenciak [4] proved that for n 7, capt(n) = n 4, and gave a characterization of the CT-maximal graphs. More recently, Kinnersley [5] has studied upper bounds on capture time on graphs where more than one cop is required to catch the robber. Gavenciak s proof relies on a detailed analysis of the conceivable cop and robber strategies, and uses a computer search at one step. In Theorem 4., we use corner rank to give a different proof, one which avoids a computer search. Ourapproachtotheproofsistoassociatecop-wingraphswithvectors, wherebyavector we simply mean a finite list of integers. The corner ranking procedure assigns each vertex in a cop-win graph an integer, so in Section we define the rank cardinality vector of a cop-win graph as the vector whose i th entry is the number of vertices of corner rank i. Since the length of the vector is the corner rank of the graph, which determines capture time, we can characterize the CT-maximal graphs by determining which vectors are realizable, i.e. which vectors are the rank cardinality vector for some cop-win graph. Thus the fundamental issue in our paper becomes determining which vectors are realizable and which are not. In Section we determine enough about the realizability of vectors to prove Theorem 4.. In Section 5 we turn to the general question of realizability; we motivate this question by showing how understanding realizability helps us understand the structure of the following interesting class of graphs. Definition.2. Let G t n be the set of cop-win graphs with n vertices and capture time t. In Section 5 we prove more about realizability, enough to allow us to characterize G n 5 Our two main theorems are Theorem 4. and Theorem 5.2, characterizing Gn n 4 and G n 5 2 n. n,

3 respectively. In Section 6 we suggest characterizing realizable vectors as a direction for future work, and state some preliminary results. 2 Corner Ranking In this section we state the definitions and theorems about corner rank that are necessary for this paper. For a full development including proofs and examples, see [7]. In this paper all graphs are finite and non-empty, i.e. they have at least one vertex; all numbers are integers. We follow a typical Cops and Robbers convention by assuming that all graphs are reflexive, that is all graphs have a loop at every vertex so that a vertex is always adjacent to itself; we will never draw or mention such edges. This assumption simplifies much of the following discussion (for example when we define homomorphism) while leaving the game play unchanged. For a graph G, V(G) refers to the vertices of G and E(G) refers to the edges of G. If G is a graph and X is a vertex or set of vertices in G, then by G X we mean the subgraph of G induced by V(G)\X. We say that a vertex u dominates a set of vertices X if u is adjacent to every vertex in X. Given a vertex v in a graph, by N[v], the closed neighborhood of v, we mean the set of vertices adjacent to v. Since all graphs in this paper are reflexive, v N[v]. For distinct vertices v and w, if N[v] N[w] then we say that v is a corner and that w corners v; if N[v] N[w], we say that v is a strict corner and that w strictly corners v; if N[v] = N[w], we call v and w twins. A cop-win ordering of a graph(also called a dismantling ordering)[6, 0] is produced by removing one corner at a time, until all the vertices have been removed or there is no corner to remove. As a small but significant modification of the cop-win ordering, rather than removing one corner at a time, we remove all the current strict corners simultaneously, assigning them a number we call the corner rank. Definition 2. (Corner Ranking Procedure). For any graph G, we define a corresponding corner rank function, cr, which maps each vertex of G to a positive integer or. We also define a sequence of associated graphs G [],...,G [α]. 0. Initialize G [] = G, and k =.. If G [k] is a clique, then: - Let cr(x) = k for all x G [k]. - Then stop. 2. Else if G [k] is not a clique and has no strict corners, then:. Else: - Let cr(x) = for all x G [k]. - Then stop.

4 b 2 a 4 4 a 2 a a 2 c c b b 2 b c c 2 d d Figure : Two representations of the graph H 7. - Let V be the set of strict corners in G [k]. - For all x V, let cr(x) = k. - Let G [k+] = G [k] V. - Increment k by and return to Step. Define the corner rank of G, denoted cr(g), to be the same as the vertex of G with largest corner rank; we understand to be larger than all integers. In [7], we show that the corner ranking procedure is well-defined. As an example, we apply the corner ranking procedure to the graph H 7 in Figure ; this graph was introduced in [2], more typically drawn like the graph on the left. The corner ranking procedure begins by assigning the strict corner d rank. After d is removed, c and c 2 are strict corners, and are thus assigned corner rank 2. Likewise, b and b 2 are assigned corner rank. After b and b 2 are removed, the remaining vertices, a and a 2, form a clique and so are assigned corner rank 4; thus the corner rank of the graph is 4. Convention. In all the figures, when a vertex w has rank k and is strictly cornered in G [k] by a vertex v of higher rank, we draw the edge vw with a thick line. Also, the number drawn inside a vertex indicates its corner rank. As another example, consider Figure 2. While cr(x) = and cr(y) = 2, once x and y have been removed there are no strict corners, and what remains is not a clique, so the other 5 vertices have corner rank ; thus the graph has corner rank. Convention. Since graphs with corner rank are a cliques, and thus a trivial case, we will assume all graphs have corner rank at least 2 For cop-win graphs, the capture time depends on a structural property of the highest ranked vertices. 4

5 v 4 v 5 v y x 2 v v 2 Figure 2: The corner ranking of a non cop-win graph. Definition 2.2. Suppose G is a graph with a finite corner rank α. We say that G is a -top graph if one of the two equivalent conditions holds: - Some vertex of corner rank α dominates V(G [α ] ). - Every vertex of corner rank α dominates V(G [α ] ). Otherwise we say G is a 0-top graph. We now state the main result (Theorem 6.) of [7], which relates the corner rank of a graph to its capture time; for a graph G, we let capt(g) = the capture time of G (note that the capt function is overloaded so that it makes sense for a graph or an integer as input; see Definition.). Theorem 2. ([7], Theorem 6.). A graph is cop-win if and only if it has finite corner rank. Furthermore, for an r-top graph G, capt(g) = cr(g) r. For example, the graph H 7 in Figure is -top with corner rank 4, so it is cop-win with capture time 4 =. In Figure 2, the graph is not cop-win and has corner rank. In [7], Lemma 2., we prove the following useful technical property; we will use this property so often that we will refer to it by a name: Upward Cornering. Lemma 2.4 (Upward Cornering). If a vertex v has corner rank k in a graph G of rank larger than k, then v is strictly cornered in G [k] by a vertex of higher rank. Rank Cardinality Vectors and Realizability Convention. For the remainder of the paper, we assume G is a cop-win graph with finite corner rank α 2. For k α, let V (G) k denote the set of vertices of rank k in G; we just write V k if the graph is apparent from context. By the term vector, we mean a finite list of positive integers. 5

6 A vector x = (x α,x α,...,x ) has length α and sum (x α + + x ). We use typical conventions for representing vectors, writing, for example, 2,...,2 to mean a list of some number of 2 s (at least one). In this section, we introduce the rank cardinality vector of a cop-win graph, which is a vector whose entries correspond to the number of vertices of each rank. Definition.. The rank cardinality vector of a graph G is the vector (x α,x α,...,x ), where x k = V k. Definition.2. A vector x = (x α,x α,...,x ) is realizable if it is the rank cardinality vector of some cop-win graph G. We say that G realizes x, or that x is realized by G. For r {0,}, x is r-realizable if there is an r-top graph H that realizes it. We say that H r-realizes x, or that x is r-realized by H. For example, the graph H 7 in Figure realizes (2,2,2,), so since H 7 is a -top graph, (2,2,2,) is -realizable. We will see that some vectors are not realizable. Since an r-realizable vector with sum n and length α corresponds to an r-top graph on n vertices with capture time α r, to understand Gn α r, to determine capt(n), and to answer related questions, the following question is of fundamental interest. Question.. For r {0, }, which vectors are r-realizable? In this section, we answer this question to the extent necessary to give a proof of Theorem 4.. In Section 5, we develop this question further and explore the general issue of realizability.. Augmentations, Initial Segments, and Extensions We introduce three ways to alter a realizable vector to obtain another realizable vector: taking an augmentation, initial segment, or standard extension. Definition.4. Consider a vector (x α,...,x ). - If the vector (y α,...,y ) has the property that x i y i for all i α, we say that (y α,...,y ) is an augmentation of (x α,...,x ). - For k, any vector of the form (x α,...,x k ) is called an initial segment of (x α,...,x ). - Any vector of the form (x α,...,x,z,z 2,...,z l ) is called an extension of (x α,...,x ). If z i = x for all i l, it is called a standard extension. - For all the notions (augmentation, initial segment, extension, and standard extension), we include the trivial case in which the vector is unchanged. - We say that x y if y is an augmentation (possibly trivial) of a standard extension (possibly trivial) of x. 6

7 Forexample, astandardextensionof(,2,2)is(,2,2,2,2)andanaugmentationof(,2,2,2,2) is (5,2,6,2,), so (,2,2) (5,2,6,2,). Lemma.5. If a vector is r-realizable, then so is any augmentation of it. Proof. It suffices to show that if x = (x α,...,x ) is r-realizable, then so is y = (y α,...,y ), where for some k, y k = x k +, and for j k, y j = x j. Consider a graph G which r-realizes x. Choose a vertex v V k, and let G be the graph obtained by adding a twin of v to G. Then G r-realizes the vector y. Lemma.6. If a vector is r-realizable, then so is any initial segment. Proof. If G r-realizes the vector (x α,...,x ), then the initial segment (x α,...,x k ) is realized by G [k]. Lemma.7. Suppose x = (x α,...,x ) and y = (x α,...,x,y k,...,y ) is a standard extension. If x is r-realizable then so is y. Moreover, if H realizes x, then there is a graph G realizing y such that G [k+] = H. Proof. It suffices toshow that if x = (x α,...,x ) isr-realizedby H, then (x α,...,x,x ) is r- realized by some G where G [2] = H. Suppose H r-realizes x with rank vertices v,...,v x. Let G be the graph obtained by adding the following to H: vertices w,...,w x and edges v w,...,v x w x. Then the vertices w,...,w x are the only strict corners in G, the rank cardinality vector of G is (x α,...,x,x ), and G [2] = H. From Lemmas.5 and.7, we conclude the following. Corollary.8. For two vectors x and y where x y, if x is r-realizable, then y is r- realizable. As a special case, note that if x = (x α,...,x ) is r-realizable and x =, then any extension of x is r-realizable. We will often use the contrapositive form of Corollary.8: If x y, and y is not r-realizable, then x is not r-realizable. For example, in Corollary.24 we show that for any k, the vector (,,k,) is not realizable, which also implies that any vector of the form (,2,k,) is not realizable..2 Projections and Path Contraction Lemma 2.4 allows us to define what we call projection functions. Again, we quote the relevant content from [7], though here we assume the graphs have finite corner rank, thus simplifying the definitions. We write f : H G to mean that f is a function whose domain is the non-empty subsets of V(H) and whose codomain is the non-empty subsets of V(G). Definition.9. Suppose G is a graph with finite corner rank α. We define the functions f,...,f α and F,...,F α,f α, where f k : G [k] G [k+] and F k : G G [k]. 7

8 - For a single vertex u V(G [k] ), define: { {u} f k ({u}) = the set of vertices in G [k+] that strictly corner u in G [k] if cr(u) > k otherwise - f k ({u,...,u t }) = i t f k ({u i }) - Let F : G G be the identity function - For < k α, let F k = f k f For a function h whose domain is sets of vertices, we adopt the usual convention that h(u) = h({u}) for a single vertex u. We say v is a k-projection (or simply a projection) of w if v F k (w). The key property of the projection functions, proved in [7], is that they are homomorphisms. Definition.0. Given two graphs H and G, and a function h : V(H) V(G), we say that h is a homomorphism if for vertices u,v V(H) and vertices u h(u),v h(v): u is adjacent to v implies u is adjacent to v. Lemma.. [7] Given a graph G, its associated functions f k and F k are homomorphisms. Definition.2. Suppose G is a graph, and H is a subgraph of G, where the vertices of H are {v,...,v k }. A k-projection of H is a graph induced by a set of vertices {v,...,v k }, where v i is a k-projection of v i. We will often refer to the k-projections of paths. We denote a path with m vertices by P m, say it has length m, and represent it by the vertices (v,...,v m ) where for i < m, v i is adjacent to v i+. Lemma. follows directly from Lemma.. Lemma.. Let G be a graph with rank α and P = (v,v 2,...,v m ) be a path in G. For all k α, every k-projection of P contains a path (possibly a single vertex) from v F k(v ) to v m F k(v m ) whose length is at most m. The following corollary will be used so often, we will refer to applications of it by the name Path Contraction. Corollary.4 (Path Contraction). If v and w are vertices in G of rank k where the shortest path from v to w in G [k] has length m, then there is no path from v to w in G of length less than m. Corollary.4 will be used as a tool to show many configurations are impossible. For example, if v and w are nonadjacent vertices of rank k without a common neighbor of rank k or higher, they cannot have a common neighbor at all. 8

9 . Vectors: Realizable and Not Realizable We now prove a number of results about realizability: some showing that a particular kind of vector is realizable, some showing that a particular kind of vector is not realizable, and some placing restrictions on the structure of graphs realizing particular kinds of vectors. Lemma.5. Suppose v is a vertex of rank k >. Then for every vertex w that strictly corners v in G [k], v must have a neighbor of rank k that is not adjacent to w. Proof. If not, then there is a vertex w that strictly corners v in G [k ], contradicting the assumption that v has rank k. Corollary.6. In a graph with rank α, every vertex of rank k > has at least one neighbor of rank k. In particular, if there is exactly one vertex v of rank k, for some k < α, then v is adjacent to all the vertices of rank k +. Lemma.7. In a graph with rank α, no vertex of rank α can dominate V α. Proof. Suppose somevertex bofrankα dominatesv α. ByUpward Cornering, letabea vertex of rank α that strictly corners b in G [α ]. Then a must also dominate V α, making G -top. In a -top graph, every vertex of rank α dominates V α, and so b is adjacent to every vertex of rank α. Thus b is adjacent to every vertex in G [α ], contradicting the assumption that a strictly corners b in G [α ]. Corollary.8. No vector (x α,...,x ) with x α = is realizable. Lemma.9. No vector (x α,...,x ) with x α 2 = is realizable. Proof. Suppose G is a graph realizing (x α,x α,...,x ), where x α 2 =, and c is the unique vertex of rank α 2. By Corollary.6, V α N[c]. By Upward Cornering, some vertex x of rank at least α strictly corners c in G [α 2]. If x V α, then x dominates V α, which contradicts Lemma.7. If x V α, then x is adjacent to every vertex in G [α 2]. Thus G [α 2] has rank at most 2, which contradicts the assumption that G [α 2] has rank. While the set of realizable vectors includes vectors that are not 0-realizable, the set of realizable vectors is in fact the same as the set of -realizable vectors. Lemma.20. Every realizable vector is -realizable. Proof. Suppose x = (x α,...,x ) is a realizable vector, realized by G. If x α =, then G must be -top so x is -realizable (though not 0-realizable). Suppose x α >. By Corollary.8 and Lemma.9: x α,x α 2 >. By Lemma.5, it suffices to show that we can -realize (2), (2,2), (2,2,2), and any vector of the form (2,2,2,,...,). Since all of these vectors are initial segments or standard extensions of (2,2,2,), which is realized by the -top graph H 7 (see Figure ), they are all -realizable. Theorem.2. 9

10 (i) The vector (,2,...,2) of length α is uniquely realized by P 2α. (ii) The vector (,2,...,2,) is not realizable. (iii) The vector (2,...,2) of length α is uniquely 0-realized by P 2α. (iv) The vector (2,...,2,) is not 0-realizable. Proof. Proof of (i): The statement is true by inspection for α =,2. It is clear that P 2α realizes (, 2,..., 2); we proceed by induction, with base case α =, to show the uniqueness. Base case (α = ): Consider any graph G realizing (,2,2); suppose V = {a}, V 2 = {b,b 2 }, and V = {c,c 2 }. The vector (,2) is uniquely realized by P, so b andb 2 arenotadjacent. Iftheyarebothadjacenttoc, thenbyupward Cornering a must strictly corner c. In order for b and b 2 to not be strictly cornered by a in G, they must each be adjacent to c 2 and a must not. But then no vertex of rank 2 or strictly corners c 2, contradicting Upward Cornering. Thus each vertex of rank 2 has a unique neighbor of rank, so we assume that b c,b 2 c 2 E(G), while b c 2,b 2 c / E(G). By Lemma.5, a cannot be adjacent to either c or c 2, and thus for i =,2, by Lemma 2.4, c i must be strictly cornered by b i. Thus c c 2 / E(G), and G = P 5. Inductive step: Now consider a graph G with rank α 4 realizing the vector (,2,...,2). By the inductive hypothesis, G [2] = P 2α = (v,v 2,...,v 2α ). Since α 4, the shortest path in G [2] between v and v 2α (which are the two rank 2 vertices in G) has length at least four. Let y and z be the two rank vertices in G (see Figure ). By Lemma.5, v and v 2α must each be adjacent to some rank vertex. However, by Path Contraction, v and v 2α cannot both be adjacent to the same rank vertex in G, and furthermore, y and z cannot be adjacent, or else there is a path of length 2 or between v and v 2α in G. Thus without loss of generality, assume yv,zv 2α E(G) and zv,yv 2α E(G). To show that G = P 2α we just need to rule out edges of the form yv i, where v i has rank at least (an analogous discussion holds of z). Suppose there is an edge yv i E(G) where v i has rank at least. Then the vertex that strictly corners y in G is not v, but must be adjacent to v, and so must be v 2. But in this case v 2 strictly corners v in G, contradicting the assumption that v has rank 2. So no edges from higher rank vertices to y or z are possible, and G = P 2α. Proof of (ii): Corollary.8 and Lemma.9 imply that (,) and (,2,) are not realizable. For α 4, if G is a graph realizing (x α,...,x ) with x α = x = and x k = 2 for 2 k < α, then by (i), G [2] = P 2α and the two rank 2 vertices u and v in G have distance 2α 4 4 in G [2]. If there were one vertex of rank, then by 0

11 v 2 v y v 2α 4 v 2α z Figure : The unique graph realizing (,2,...,2) is P 2α. Corollary.6 the rank vertex is adjacent to both u and v, yielding a length 2 path from u to v, contradicting Path Contraction. Proof of (iii): This proof is almost the same as the proof of (i), but now with a base case stating that (2,2,2) is uniquely 0-realized by P 6 ; the proof of the base case is a similar technical proof to that of the base case for (,2,2). Proof of (iv): This proof is the same as the proof of (ii), using (iii) instead of (i). We now turn our attention to graphswith rank 4, starting with a simple technical lemma. Lemma.22. If a graph realizes (a,b,c,) then there is a vertex of rank or 4 that dominates the rank 2 vertices. Proof. Let G be the graph and let d be the lone vertex in V. By Corollary.6, V 2 N[d]. By Upward Cornering, some vertex x of rank greater than must strictly corner d, so V 2 N[x]. If x V 2, then by Upward Cornering let y be a vertex of rank at least that strictly corners x in G [2], otherwise let y = x. In either case, we have a vertex y in either V or V 4 such that V 2 N[y]. Theorem.2. Suppose a graph realizes (,m,k,). Then the subgraph induced by the rank vertices is connected. Proof. Let G be the graph and let H be the subgraph induced by the rank vertices. Assume for the sake of contradiction that the claim is false. Suppose a is the rank 4 vertex, two components of H have vertex sets B and B 2, and for i =,2, b i B i. By Lemma.5, there must be a rank 2 vertex c adjacent to b but not to a. Since b is only adjacent to rank vertices in B, by Upward Cornering, c must be strictly cornered in G [2] by a vertex in B and thus c is only adjacent to rank vertices in B. Similarly, there is a rank 2 vertex c 2 that is adjacent to b 2, but not to a; likewise, c 2 is only adjacent to rank vertices in B 2. If c and c 2 are adjacent or have a common neighbor c of rank 2 then the vertex of higher rank (which we have by Upward Cornering) that strictly corners c (or c if c and c 2 are

12 adjacent) in G [2] would have to be adjacent to both c and c 2. However, no such higher rank vertex exists since it would have to be in both B and B 2, but these sets are disjoint. Thus c and c 2 are at distance at least three in G [2], and by Path Contraction, they cannot both be adjacent to the single rank vertex, contradicting Corollary.6. Since the graph induced by the rank vertices of any graph realizing (,) is not connected, Lemma.2 implies the following corollary. Corollary.24. For all k, the vector (,,k,) is not realizable. Theorem.25. (i) For k, (2,4,k,) is not 0-realizable. (ii) (2, 5, 2, ) is not 0-realizable. Proof. The proofs of (i) and (ii) are similar, with only some differences at the end. Consider for the sake of contradiction a graph G that 0-realizes (2,m,k,) or (2,5,2,). Since G is a 0-top graph and V 4 = {a,a 2 } has only two vertices, there are rank vertices b and b 2 such that a b,a 2 b 2 E(G) and a b 2,a 2 b E(G). For i =,2, a i must strictly corner b i and every rank neighbor of b i in G [] ; we will use this point throughout the proof. Since no rank 4 vertex is adjacent to both b and b 2, they can share no common neighbors in G [] (since no rank 4 vertex could corner such a vertex in G [] ), and by Path Contraction, b and b 2 must be at distance at least in G. For i =,2, let c i be a rank 2 vertex adjacent to b i but not a i, which must exist by Lemma.5. Since the distance between b and b 2 is at least, c and c 2 must be distinct vertices, and b c 2,b 2 c E(G). Since no vertex of rank 4 dominates V 2, by Lemma.22, there is a vertex b of rank that dominates V 2, and b is not b or b 2. Without loss of generality suppose a 2 corners b in G []. Now consider what corners c in G [2] : neither a i, not b 2 because it is not adjacent to c, and neither b nor b since that would force b and b to be neighbors and would imply a 2 is adjacent to b, a contradiction. So a fourth distinct rank vertex b 4 must corner c in G [2], and thus b 4 must be adjacent to both b and b. To finish the proof for (i): Now consider what vertex of rank at least strictly corners c 2 in G [2]. Since the distance from b to b 2 is at least, neither of b or b 4 can be adjacent to b 2 and thus neither of these vertices can corner c 2. Neither vertex of rank 4 works since a is not adjacent to b 2 and a 2 is not adjacent to c 2. So b 2 or b strictly corners c 2 in G [2], and are thus adjacent to each other. But now b 2 is strictly cornered by b in G [2], since they have the same neighbors in G [2], except that b is adjacent to c and b 4, while b 2 is not. To finish the proof for (ii): Since a 2 is not adjacent to b, a must corner b 4 in G [], so in particular a and b 4 are adjacent. Since b is not strictly cornered by b 4 in G [2], it must be adjacent to the fifth rank vertex b 5, while b 4 and b 5 are not adjacent. Since b 4 corners c, b 5 is not adjacent to c, so by Lemma.5, b 5 must be adjacent to c 2. Since a must strictly corner b 5 in G [], b 5 is not adjacent to b 2, and thus b is the only vertex that can strictly corner c 2 in G [2]. But then b is adjacent to the rank vertices b 2, b 4, and b 5, and thus also a. Thus in G [], b has at least the neighbors that a 2 has, contradicting the fact that a 2 strictly corners b in G []. 2

13 Theorem.26. For any m,k, (m,2,k,) is not 0-realizable. Proof. For the sake of contradiction, suppose G 0-realizes (m,2,k,). Let V = {b,b 2 } and note that every rank 4 vertex is adjacent to exactly one of these two vertices. Thus b b 2 E(G), and these two vertices are at distance in G [] and hence in G. Thus by Path Contraction they share no rank two neighbors. By Lemma.22, there is a vertex x of rank or 4 that dominates V 2. Since b and b 2 must both have rank 2 neighbors but can t have any in common, neither of these vertices can be x. Thus x must be a rank 4 vertex. But if x is adjacent to b i, then it strictly corners b i in G [2], contradicting the assumption that b i has rank. Recall the graph H 7 from Figure. Lemma.27. The vector (2,2,2,) is uniquely realized by H 7. Proof. Let G be a graph that realizes (2,2,2,), with V 4 = {a,a 2 }, V = {b,b 2 }, V 2 = {c,c 2 }, and V = {d}. Theorem.2 implies that (2,2,2,) is not 0-realizable. Thus G [] must contain the edges a a 2, a b, a 2 b 2, a b 2, a 2 b, and since G [] is not a clique, there is not an edge b b 2. By Corollary.6, each of b and b 2 must be adjacent to a vertex of rank 2, and Lemma.22 implies that some vertex x of rank or 4 dominates V 2. If x were some a i, then x would strictly corner each rank vertex in G [2], a contradiction. Thus without loss of generality we may assume b dominates V 2 and both b and b 2 are adjacent to c. Then only a vertex from V 4 can strictly corner c in G [2] ; without loss of generality, suppose a 2 is this vertex, so in particular, a 2 is adjacent to c. Since a 2 is not a dominating vertex in G [2], it cannot be adjacent to c 2 and thus c and c 2 cannot be adjacent. For a 2 not to strictly corner b 2 or a in G [2], each of these vertices must be adjacent to c 2, and a cannot be adjacent to c or else it dominates G [2]. Thus G [2] = (H 7 ) [2]. By Corollary.6, the rank vertex d is adjacent to both c and c 2, which means it can only be strictly cornered by some b i, without loss of generality, b. Since the rank vertices are not adjacent, d cannot be adjacent to b 2. Finally, by Lemma.5, d cannot be adjacent to any rank 4 vertex. Thus G is H 7. 4 A Characterization of G n 4 n, the CT-Maximal Graphs We can now characterize the rank cardinality vectors of all the CT-maximal graphs. The following definition will be used to classify the CT-maximal graphs having at least seven vertices. Definition 4.. For k 0, define H 7 +k to be a set of graphs that realize the length 4 + k vector (2,2,2,,...,). Let H 7 + be k 0 H+k 7. For example, Lemma.27 implies that H 7 +0 = {H 7 }. Figure 4 displays some of the graphs in H 7 +. By Lemma.7, any standard extension of (2,2,2,) is realizable, so for each k, H 7 +k is non-empty. In [4], M is defined to be the set of CT-maximal graphs. We will see

14 Figure 4: Some graphs in H + 7. (in Theorem 4.) that for n 9, H + 7 is the same as M. In Theorem 2 of [4] a nice, but somewhat involved characterization of M is given (stated to be true for n 8, but actually true for n 9). Our result gives a simpler characterization (for n 9): A graph is in M exactly when it realizes (2,2,2,,...,). In the process of characterizing M, Gavenciak [4] derives various properties of the graphs in M; these properties follow almost immediately from our characterization of M by H + 7, summarized in the next theorem. Theorem 4.2. Suppose G is a graph on n vertices in H + 7. Then (i) G [α ] is H 7. (ii) G is -top. (iii) capt(g) = n 4. Proof. Property (i) follows from Lemma.27. Property (ii) follows from the fact that H 7 is -top. For Property (iii), note that G has rank n and is -top. Thus by Theorem 2., G has capture time (n ) = n 4. The next theorem restates the main results of [4], with an alternative proof that does not use a computer search. 4

15 Theorem 4.. For n 7, capt(n) = n 4, and for graphs on at least 9 vertices, the CTmaximal graphs are exactly the graphs in H 7 +. Furthermore, in Table, we describe capt(n) and the CT-maximal graphs for n 8. n capt(n) CT-Maximal Graphs with n vertices 0 P 2 P 2 P, K 4 2 P P 5 and the 0-top graphs realizing (2,) and (,2) 6 P 6 7 P 7, H 7, and the 0-top graphs realizing (2,2,), (2,,2), (,2,2) 8 4 P 8 and any graph in H 7 + Table : CT-Maximal graphs with at most 8 vertices and their capture time Proof. That capt(n) = n 4 (for n 7), follows immediately from Theorem 4.2 and the rest of this theorem. We begin with the case of n 8, considering vectors realized by P n. By Theorem.2, when n is even, P n is the unique 0-top graph realizing the length n/2 vector (2,...,2), and when n is odd, P n is the unique -top graph realizing the length n/2 vector (,2,...,2). Thus when n is even, graphs whose rank cardinality vector has length less than n/2 cannot be CT-maximal, and when n is odd, graphs whose rank cardinality vector is less than n/2 cannot be CT-maximal. Based on this observation, Table 2 lists all vectors with sum n 8 that could possibly be the rank cardinality vector of some CT-maximal graph; by Corollary.8 and Lemma.9, we exclude the vectors whose second or third entry is. Note that the first vector (in bold) is the rank cardinality vector for the corresponding path P n. n Vectors () 2 (2) (,2), () 4 (2,2), (,) 5 (,2,2), (,4), (,2), (2,) 6 (2,2,2), (,,2), (,2,), (,2,2,) 7 (,2,2,2), (2,2,2,), (2,2,), (2,,2), (,2,2), (,2,2,,), (,,2,), (,2,,) 8 (2,2,2,2), (2,2,2,,) Table 2: Vectors with sum n 8 and length at least n/2. To prove the theorem for n 8, it suffices to show that each vector is either: ) not realizable, 2) has capture time less than that of P n, or ) is accounted for in Table. We 5

16 2 Figure 5: The unique graph 0-realizing (, 2) and the three graphs 0-realizing (2, ). proceed by cases on the values of n 8, employing Theorem 2. and using the immediate fact that if the first entry is, then a graph that realizes the vector must be -top. Also, recall the remarks in the paragraph directly before Theorem.2, where we discuss the issue of uniquely realizing small vectors and why we omit some proofs. At various points in this proof all we need to show is that some vector is realizable; in some of those cases, as an interesting tangent, we claim that the vector is uniquely realized, or we produce all the graphs realizing the vector. - For n =,2, all the vectors listed in Table 2 have corresponding graphs listed in Table. - For n = 4, a graph realizing (,) has capture time < 2, so it is not CT-maximal. - For n = 5, besides (,2,2), the vectors listed in Table 2 have length less than, so they can only have capture time 2 if they are 0-top, which means we also get as CT-maximal graphs the unique graph 0-realizing (, 2) and the three graphs 0-realizing (2, ). (See Figure 5.) - For n = 6, the only vector, besides (2,2,2), corresponding to a capture time of or greater is (,2,2,), but that vector is not realizable, by Theorem.2. - For n = 7, the vector (2,2,2,) is uniquely realized by H 7, using Lemma.27. To achieve the required capture time of, we can also take one of the five graphs 0- realizing (2,2,) or one of the unique graphs 0-realizing (2,,2), or (,2,2). (See Figure 6.) The rest of the vectors are not realizable: (,2,2,,) is not realizable by Theorem.2, and (,,2,) and (,2,,) are not realizable by Corollary For n = 8, by definition, the vector (2,2,2,,) is only realized by graphs from H + 7. Now we consider n 9. We show that H +(n 7) 7 contains all the CT-maximal graphs. For H +(n 7) 7 not to contain all the CT-maximal graphs we would need a realizable vector x = (x α,...,x ) besides (2,2,2,,,...,) with one of the following properties. - Type 0: α n 4 and x is 0-realizable. - Type : α n and x is -realizable. 6

17 2 Figure 6: Top: The five graphs 0-realizing (2, 2, ). Bottom: The unique graphs 0-realizing (2,,2) and (,2,2). We show that no such vectors are realizable. Keep in mind that in both cases x α and x α 2 must be at least 2 by Corollary.8 and Lemma.9. We rule out the type 0 vectors. Let y = (y α,...,y ) be the vector (2,2,2,,...,). Being 0-realizable, x α 2. Since α n 4, such an x would be an augmentation of y where all entries of x are the same as the entries of y with the possible exception of one entry of y, which is one larger than its corresponding entry in x. No matter where the is added, or if nothing is added, one of the following vectors must be an initial segment of x: (,2,2,), (2,,2,), (2,2,,), (2,2,2,) or (2,2,2,2,). The first and third vectors are not 0-realizable by Theorem.26, and the second is not 0-realizable by Theorem.25; the last two vectors are not 0-realizable by Theorem.2. Now we rule out the type vectors. Let y = (y α,...,y ) be the vector (,2,2,,...,). Since α n, such an x would be an augmentation of y where all entries of x are the same as the entries of y with the possible exception of one entry of y, which is one larger than its corresponding entry in x. The value cannot be added to y α since that would mean G is in H + 7. No matter where else is added, or if nothing is added, one of the following vectors must be an initial segment of x: (,2,2,), (,2,2,2,), (,,2,), (,2,,). By Theorem.2 the first two vectors are not realizable, and by Corollary.24, the last two vectors are not realizable. 5 A Characterization of G n n 5 Before we prove our second main result, Theorem 5.2, we need the following lemma. 7

18 Lemma 5.. Let x α,...,x have the property that x j = for some j >, and x i = 2 for all i j. Then (i) There is exactly one graph that realizes (,x α,...,x ). (ii) (,x α,...,x,) is not realizable. (iii) There is exactly one graph that 0-realizes (x α,...,x ). (iv) (x α,...,x,) is not 0-realizable. Proof. Proof of (i): We first suppose we have a vector x of the form (,,2) or (,2,...,2,,2), and we will show that it is uniquely realized, so we let G be this unique graph. If x is (,,2) or (,2,,2), we will show that the corresponding graph G is drawn in Figure 7. Otherwise, we are considering an x of length at least 5, of the form (,2,...,2,,2); in this case the corresponding graph G is partially drawn on the right side of Figure 7: its bottom four ranks are drawn; also there are no edges between V(G [5] ) and any vertex of rank less than 4. Once we have shown that such a vector x corresponds to such a unique graph G, we can quickly obtain the uniqueness claim for any vector which is a standard extension of x. Considering any such standard extension of x, using the properties of G, and key facts like Path Contraction, we can see that any such standard extension is only realized by attaching an appropriate length path to each of the rank vertices of G. The bulk of the proof now consists in showing that vectors of the form x are uniquely realized in the manner described. We first deal with the cases of (,,2) and (,2,,2). It is a simple exercise to see there is only one graph that realizes (,,2) (see Figure 7). Now we show that there is only one graph that realizes (,2,,2). Suppose G realizes (,2,,2), with V 4 = {a}, V = {b,b 2 }, V 2 = {c,c 2,c } and V = {d,d 2 }. There are 4 graphs realizing (,2,) (note to the reader: in finding them, note that two have an edge between a and V, and two do not). In each of the 4 graphs we can assume without loss of generality that b is adjacent only to a and c, and c has degree. Thus c is at distance at least from any other rank 2 vertex of G, and in any realization of (,2,,2), c must be adjacent to a vertex d that is not adjacent to b, c 2 or c. This implies c must strictly corner d. The vertex d 2 must be adjacent to c 2 and c, and the only way to fill in the rest of the edges leads to Figure 7 (to help see this, note that neither b 2 nor a can strictly corner d 2 ). WenowconsiderthecasewhereG isagraphofrankatleast5thatrealizes(,2,...,2,,2). Let V 4 = {a,a 2 }, V = {b,b 2 }, V 2 = {c,c 2,c } and V = {d,d 2 }; we will show, without loss of generality, that the graph induced by these vertices of rank 4 and less, is pictured in Figure 7, on the right side, and that there are no edges between G [5] and the vertices of rank less than 4. By Theorem.2, ( ) G [] is uniquely realized as a path. 8

19 2 a 4 b b 2 c 2 c 2 c d d 2 a 4 4 a 2 b b 2 c 2 c 2 c d d 2 Figure 7: The unique graphs realizing (,,2) and (,2,,2), and the four lowest ranks of the unique graph realizing (,2,...,2,2,,2). By ( ), and without loss of generality, a is adjacent to b, a 2 is adjacent to b 2, and the distance between b and b 2 in G [] is at least 4. Thus by Path Contraction, the distance between b and b 2 in G is at least 4. Thus b and b 2 cannot share any neighbors of rank 2, so without loss of generality we can assume b is adjacent to c but not c 2 and b 2 is adjacent to c 2, but not c. We now make an observation: If the only rank 2 neighbor of b i is c i, then b i must strictly corner c i in G [2]. Consider why the observation is true. Since c i is adjacent to b i, by ( ), the only vertices that could strictly corner c i in G [2] are a i and b i. If a i strictly cornered c i in G [2] then it would also strictly corner b i in G [2], which cannot happen, so b i must strictly corner c i in G [2]. So the observation is true. As mentioned above, at most one of b or b 2 can be adjacent to c, so for some i, the only rank 2 neighbor of b i is c i. Thus the shortest path in G [2] between c and c 2 must include b i and a i, so by Path Contraction, c and c 2 cannot be adjacent, nor adjacent to the same vertex. Thus without loss of generality, d is adjacent to c and not c 2, and d 2 is adjacent to c 2 and not c. Now c must be adjacent to one of the rank vertices, without loss of generality d 2. Since c 2 and c are at distance at most 2 in G, by Path Contraction, in G [2] they are at distance at most 2, from which we can conclude that there is a vertex x in G [] that is adjacent to both c 2 and c (note that if c 2 and c were adjacent, then the vertex x will be the vertex that strictly corners c in G [2] ). We show that b 2 must be adjacent to c, by assuming for contradiction that it were not. Then by the observation, b 2 must strictly corner c 2 in G [2], so a 2 is not adjacent to c 2 and so cannot be x. By assumption, x is not b 2. Since b 2 strictly corners c 2 in G [2], b 2 has to be adjacent to x violating ( ). So we have that b 2 is adjacent to both c 2 and c. Thus, just as we argued that c 2 is not adjacent to d, so c is not adjacent to d. Now, by ( ), only a 2 or b 2 can strictly corner either c 2 or c in G [2], but since a 2 cannot be adjacent to both c 2 and c, b 2 must strictly corner at least one of c 2 and c ; without loss of generality, assume b 2 strictly corners c in G [2]. Now consider what vertex y strictly corners d 2. The vertex y would have to be adjacent to at least d 2,c 2, and c. We know y a 2 since a 2 cannot be adjacent to both c 2 and c. The vertex y cannot be another vertex in G [4], 9

20 Figure 8: A graph -realizing (,4,2,) and a graph 0-realizing (,,2,). since then y would be adjacent to c and since b 2 strictly corners c in G [2], b 2 would have to be adjacent to y, violating ( ). The vertex y can also not be b 2 since then b 2 would in fact strictly corner c in G. Thus d 2 is strictly cornered by one of c 2 or c, meaning that c 2 is adjacent to c. Viewing Figure 7, we have shown that all the displayed edges must be there and have ruled out most of the missing edges; we just need to rule out a few more edges. We rule out any other edges attached to c 2 by considering what could corner c 2 in G [2] : not a 2 since then a 2 would be adjacent to c 2 and c, and not any other vertex in G [4], since by ( ) it is not adjacent to b 2. So only b 2 can strictly corner c 2 in G [2], so there can be no more edges attached to c 2. We rule out an edge between d and d 2 using Path Contraction, since by the reasoning to this point we can now conclude that the distance between c and c 2 is at least 5 in G [2]. Also d 2 can have no neighbors besides c 2 and c because if it did, then nothing could strictly corner it; similarly, d can have no other neighbors besides c. Proof of (iii): The argument is the same as the one for (i), with 0-realizations of (,2), (2,,2) and (2,...,2,,2) in place of (,,2), (,2,,2), and (,2,...,2,,2). Proofs of (ii) and (iv): Assume for contradiction that we had a graph G realizing the appropriate vector. Thus G [2] is as described in parts (i) and (iii), so the two rank 2 vertices of G are at distance greater than 2 in G [2], but by Corollary.6, must both be adjacent to the rank vertex in G, contradicting Path Contraction. Theorem 5.2. A cop-win graph on n vertices has capture time n 5 if and only if one of the following conditions holds: - It -realizes a standard extension of (,4,2,). - It -realizes a vector formed by taking a standard extension of (2,2,2,) and then augmenting by adding to any single entry. - It 0-realizes a standard extension of (,,2,). 20

21 Proof. By Theorem 2. we know that any graph satisfying one of the conditions does have capture time n 5. Observing Figures and 8 we see that we can -realize (,4,2,) and (2,2,2,), and 0-realize (,,2,); thus the three classes of graphs in the statement of the theorem are non-empty. It remains to show that our three conditions have not missed any graphs. Let G be a cop-win graph on n vertices, with capture time n 5, with rank cardinality vector x = (x α,...,x ). Since n, x must have length at least 6, and at least one of the first 6 entries of x, besides x α, must be a (since otherwise Theorem 2. would imply G has capture time less than n 5). So suppose x i = and x j > for i < j < α, and note that i α by Corollary.8 and Lemma.9. Consider cases on whether G is 0-top or -top. - Case: G is 0-top. If x i is x α 5, then in order to have capture time n 5, we must have (2,2,2,2,2,) as an initial segment of x, but this vector is not 0-realizable by Theorem.2. If x i is x α 4, then in order to have capture time n 5, we have the following possible initial segments of x: (,2,2,2,), (2,,2,2,), (2,2,,2,), or (2,2,2,,). The first three vectors are not 0-realizable by Lemma 5.. We can show the vector (2,2,2,,) is not 0-realizable using Theorem.2 and Path Contraction. If x i is x α, then in order to have capture time n 5, the possible initial segments are: (,,2,), (,2,,), (2,,,), (4,2,2,), (2,4,2,), or (2,2,4,). The first vector (,,2,) is 0-realizable as required; we show that the rest are not 0-realizable. The vectors (2,,,), (2,4,2,), (2,2,4,) are not realizable by Theorem.25. The vectors (,2,,) and (4,2,2,) are not 0-realizable by Theorem Case: G is -top. If x i is x α 5, then in order to have capture time n 5, we must have (,2,2,2,2,) as an initial segment of x, but this vector is not realizable by Theorem.2. If x i is x α 4, then in order to have capture time n 5, we have the following possible initial segments of x: (,,2,2,), (,2,,2,), or (,2,2,,). By Lemma 5. the first two are not realizable. We can show the vector (,2,2,,) is not realizable using Theorem.2 and Path Contraction. If x i is x α, then in order to have capture time n 5, the possible initial segments are: (,4,2,), (,2,4,), or (,,,). The first vector is realizable as required. The other two are not realizable by Corollary Future Work The main results of our paper are structural characterizations of Gn n 4 the big open question for us is the following. and Gn n 5. Naturally 2

22 Question 6.. Find structural characterizations of G t n for all t n 4. Our approach is to give the charaterization in terms of what vectors the graphs should realize. With some terminology, we will be more specific about our approach. Definition 6.2. A vector x, of length at least 2, is r-minimal if the only r-realizable vector x, of length at least 2, is x itself. A vector is minimal if it is either 0-minimal or -minimal. For example, it follows from Theorem 4. that (2,2,2,)is -minimal, and thus, for example, (2,7,2,) and (2,2,2,,,) are not -minimal. We can restate the crux of our main results as follows (recall the ordering on vectors from Definition.4): - For n 9, Gn n 4 is the set of graphs with n vertices that -realize a vector of length n which is larger than (2,2,2,). - G n 5 n is the set of graphs with n vertices that either:. 0-realize a vector of length n 5 which is larger than (,,2,), or 2. -realize a vector of length n 4 which is larger than (,4,2,) or (2,2,2,). A general approach to characterizing some G t n is to find the appropriate minimal vectors and take the vectors that are larger of appropriate length. The key technical point then becomes determining which vectors are minimal. In other words, we can make Question. more specific: Question 6.. For r {0,}, which vectors are r-minimal? In [8], we have a collection of examples working in this direction, which we summarize here without proof. While we know the vectors in Theorems 6.4 and 6.5 are minimal, we conjecture that the vectors in Theorems 6.6 and 6.7 are minimal. Theorem 6.4. The vectors (,2), (,4,2,), and (2,2,2,) are -minimal. Theorem 6.5. The vectors (2,2), (2,5,,), (2,6,2,), and (,,2,) are 0-minimal. Theorem 6.6. The following vectors are -realizable. (,2,8,4,) (,2,4,4,4,2,2,) (,,5,4,2,) (,2,6,4,2,) (,2,4,2,4,2,2,2,) (,,4,4,2,2,) (,2,5,4,,2,) (,2,,,,,2,2,2,) (,,,,,2,2,) (,2,5,,,2,2,) Theorem 6.7. The following vectors are 0-realizable. (2,4,4,2,) (2,,,,,2,2,2,) (2,4,,4,2,2,) (,2,4,2,,2,2,) (2,4,2,4,2,2,2,) (4,2,4,2,) 22

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